II.1. Show that α ∈ ℂ is algebraic if and only if |ℚ(α)∶ℚ|$<∞$. Deduce that 𝔸, the set of algebraic numbers in ℂ, is a subfield of ℂ. Assuming that the algebraic integers, $\barℤ$, form a ring, show that the field of fractions of $\barℤ$ is 𝔸.

IfIf α is algebraic, there exists irreducible $f(x)∈ℚ[x]:f(α)=α^n+a_{n-1}α^{n-1}+⋯+a_0=0$.

[See Example 5.1] $⟨f(x)⟩$ is a maximal ideal, so $ℚ[x]/⟨f(x)⟩$ is a field, so $ℚ[α]=ℚ(α)$. By division algorithm {1, α, …, α

Since 𝔸 is a field that contains $\barℤ$, we have $𝔸⊃\operatorname{Frac}(\barℤ)$.

If α ∈ 𝔸, there is

II.2. Show that 2 is irreducible but not prime in $ℤ[\sqrt{-5}]$, and that 1 is a greatest common divisor of 2 and $1+\sqrt{-5}$. Do we have $⟨2⟩+⟨1+\sqrt{-5}⟩=ℤ[\sqrt{-5}]$? Is $2+2\sqrt{-5}$ a least common multiple of 2 and $1+\sqrt{-5}$? Does 6 have a unique factorisation into irreducibles? Does 2 have a unique factorisation into irreducibles?

If $a,b∈ℤ[\sqrt{-5}]$ satisfy $2=ab$, then $4=N(a)N(b)$ where $N(a),N(b)∈ℤ^+$, so $N(a),N(b)<5$, so $a,b∈ℤ$, so $a∼1$ or $b∼1$. So 2 is irreducible in $ℤ[\sqrt{-5}]$.$2|m+n\sqrt{-5}$ iff $2|m,n$. So $2∤(1-\sqrt{-5})$ and $2∤(1+\sqrt{-5})$.

But $2|(1-\sqrt{-5})(1+\sqrt{-5})=6$. So 2 is not prime in $ℤ[\sqrt{-5}]$.

Let $a=\gcd(2,1+\sqrt{-5})$. Since $a|2$, $N(a)|N(2)=4$. Similarly $N(a)|N(1+\sqrt{-5})=6$, thus $N(a)|\gcd(4,6)=2$. There is no element with norm 2, hence $N(a)=1⇒a∼1$.

If $1=2r+(1+\sqrt{-5})s,r=a_1+b_1\sqrt{-5},s=a_2+b_2\sqrt{-5},a_1,b_1,a_2,b_2∈ℤ$, we compute \begin{align*}1=N(2r+s+s\sqrt{-5})&=(2a_1+a_2-5b_2)^2+5(2b_1+b_2+a_2)^2\\&≡(a_2+b_2)^2+(b_2+a_2)^2\\&≡0\pmod2\end{align*}contradiction$⇒1∉⟨2⟩+⟨1+\sqrt{-5}⟩⇒⟨2⟩+⟨1+\sqrt{-5}⟩≠ℤ[\sqrt{-5}]$.

$6=2⋅3=(1+\sqrt{-5})(1-\sqrt{-5})$ is a common multiple of 2 and $1+\sqrt{-5}$

But $N(2+2\sqrt{-5})=24$, so $2+2\sqrt{-5}∤6$, so $2+2\sqrt{-5}$ is not a LCM of 2 and $1+\sqrt{-5}$.

$6=2⋅3=(1+\sqrt{-5})(1-\sqrt{-5})$. It remains to prove $2,3,1±\sqrt{-5}$ are irreducible.

Let $x=a+b\sqrt{-5}|2$ and $x≁1$ then $N(x)|N(2)=4⇒b=0⇒a^2|4⇒a=2$, so 2 is irreducible, so 2 has a unique factorisation into irreducibles.

Let $x=a+b\sqrt{-5}|3$ and $x≁1$ then $N(x)|N(3)=9$, if $b=0$ then $a^2|9⇒a=3$; if $b=1$ then $a^2+5|9⇒a=2$ but $2+\sqrt{-5}∤3$ so 3 is irreducible.

Let $x=a+b\sqrt{-5}|1±\sqrt{-5}$ and $x≁1$ then $N(x)|N(1±\sqrt{-5})=6$, if $b=0$ then $a^2|6$ contradiction; if $b=1$ then $a^2+5|6⇒a=±1$ but $1±\sqrt{-5}$ doesn't divide each other, so $1±\sqrt{-5}$ are irreducible.

So 6 has non-unique factorisation into irreducibles.

II.3. Suppose that *R* is an infinite integral domain with finitely many units in which every non-unit has an irreducible factor. By emulating Euclid’s proof that there are infinitely many primes show that *R* contains infinitely many irreducible elements. Apply this to $𝔽_2[X]$ to deduce that there are arbitrarily large fields of characteristic 2.

Since $a$ is not a unit, \(a^n≠±1\) for any \(n\).

Since \(a≠0,1 - a^{n-m}≠0\) and

Since there are only finitely many units in \(R\), there exists \(n\) such that \(1 + a^n\) is not a unit.

Since every non-unit has an irreducible factor, \(1 + a^n\) has an irreducible factor, say $p_i$.

But \(p_i | a^n\) and \(p_i | 1 + a^n\), so \(p_i | 1\), a contradiction.

Applying this to $𝔽_2[X]$, we deduce that there are infinitely many irreducible elements.

Since elements of degree <

[See Example 5.1] $⟨f(X)⟩$ is a

II.4. Show that if *R* is an integral domain in which every ideal is finitely generated then *R* has the ACCP. Assuming that the algebraic integers, $\barℤ$, form a ring, show that it has no irreducible elements and deduce that not every ideal in $\barℤ$ is finitely generated.

For non-unit $α∈\barℤ$, wlog α > 0. As $α=\sqrtα⋅\sqrtα$, we show $\sqrtα∈\barℤ$ then α is not irreducible.

$α∈\barℤ⇒∃a_i∈ℤ:α^n+a_{n-1}α^{n-1}+⋯+a_0=0$

$⇒{\sqrtα}^{2n}+a_{n-1}{\sqrtα}^{2(n-1)}+⋯+a_0=0⇒\sqrtα∈\barℤ$

Suppose every ideal in $\barℤ$ is finitely generated, by first part, $\barℤ$ has the ACCP, by Prop 3.32, every

II.5. Draw a picture to explain why for every *z* ∈ ℂ there is *w* ∈ ℤ[i] with |*z* − *w*|$≤1/\sqrt2$. Hence show that the function *f*(*a* + *b*i) ≔ |*a* + *b*i|^{2} is a Euclidean function on ℤ[i], and use this to find *z*, *w* ∈ ℤ[i] such that *d* ≔ 2*z* + (3 − 5i)*w* has $d|2$ and $d|3-5i$ in ℤ[i]. Is the function $f(a+b\sqrt{-5})≔\left|a + b\sqrt{-5}\right|^2$ a Euclidean function on $ℤ[\sqrt{-5}]$? Is the function $f(a+b\sqrt{-2})≔\left|a+b\sqrt{-2}\right|^2$ a Euclidean function on $ℤ[\sqrt{-2}]$?

If

Let $a,b∈ℤ[\sqrt{-2}]$*, we'll find $q∈ℤ[\sqrt{-2}]$ such that ${|a-bq|}<b$.

Let $ab^{-1}=c+di,\,c,d∈ℝ$. Let $c',d'$ be the closest integer to $c,d$. Let $q=c+d\sqrt{-2}$.\begin{align*}f(a-bq)&=f(b)f(ab^{-1}-q)=f(b)f\left((c-c')+(d-d')\sqrt{-2}\right)\\&=f(b)[(c-c')^2+2(d-d')^2]\\&≤f(b)[(1/2)^2+2(1/2)^2]<f(b)\end{align*}(…the rest is same as ℤ[i] proof). Hence

By II.2 $ℤ[\sqrt{-5}]$ is not UFD, so it is not ED. Alternatively, the ideal $⟨3,2+\sqrt{-5}⟩$ is not principal [Suppose $⟨3,2+\sqrt{-5}⟩=⟨a⟩$, then $a≁1$ and $a|3⇒a=±3$, but $a∤2+\sqrt{-5}$], so $ℤ[\sqrt{-5}]$ is not PID, so it is not ED.

II.6. Suppose *p* is prime in ℤ. Show that if there are *a*, *b* ∈ ℤ such that *p* = *a*^{2} + *b*^{2} then *a* + *b*i is prime in ℤ[i], and if there are no such *a* and *b* then *p* is prime in ℤ[i].

If ∃

Writing

II.7. Use the fact that every irreducible element of ℚ[*X*] is prime in ℚ[*X*] to show that every irreducible element of ℤ[*X*] is prime in ℤ[*X*].

Let

Suppose

II.8. Show that ℤ[*X*] has the ACCP, and combine this with the previous exercise to deduce that ℤ[*X*] is a unique factorisation domain.

$a_{n+1}|a_n$, since ℤ has ACCP, $∃N,∀n≥N:a_n∼a_N$

$(\deg f_n)$ is a decreasing sequence in ℕ, so $∃N'≥N,∀n≥N':\deg f_n=\deg f_{N'}⇒f_n∼f_N$

ℤ[

II.9. Suppose that *R* is an integral domain. Show that *R* is a PID if and only if there is a function *f* : *R** → ℕ_{0} with $f(a)≤f(b)$ whenever $a|b$; and such that if *a*, *b* ∈ *R** then either $b|a$, or there are *x*, *y* ∈ *R* and *r* ∈ *R** with $ax=by+r$ and $f(r)<f(b)$.

If $a∉⟨b⟩$ then there is

II.10. Suppose that *R* is an integral domain with the ACCP in which every maximal ideal is principal. Show that *R* is a PID.

Lemma.

Since $U⊂U+⟨a⟩$ and $a∈U+⟨a⟩$, we have $U⊊U+⟨a⟩$, so $U+⟨a⟩=⟨c⟩$, for some $c∈R$.\begin{array}{rll} c∈U+⟨a⟩\mmlToken{mo}[rspace=0]⟹&c=u+ra&\text{for some $u∈U$ and $r∈R$}\\ ⟹&bc=b(u+ra)=bu+r(ab)∈U&\text{since $ab∈U$}\\ ⟹&b∈\{x∈R|cx∈U\}≕V \end{array} Since $U⊂V$, and $b∈V∖U$, we have $U⊊V$, so $V=⟨d⟩$ for some $d∈R$.

For any $u∈U$, \begin{array}{rll}U⊂U+⟨a⟩=⟨c⟩\mmlToken{mo}[rspace=0]⟹&u=cy&\text{for some $y∈R$}\\ ⟹&cy∈U\\ ⟹&y∈V=⟨d⟩&\text{by definition of $V$}\\ ⟹&y = dz&\text{for some }z∈R\\ ⟹&u=c(dz)\\ ⟹&u∈⟨cd⟩\\ \end{array} Thus, $U⊂⟨cd⟩$.

Also, since $d∈V$, by definition of $V$, we get $cd∈U$, hence $⟨cd⟩⊂U$.

But then $U=⟨cd⟩$, contrary to the assumption that $U$ is not principal.

It follows that $U$ is prime. Alternate proof:

Let $I_0$ be a non-principal ideal, since $\{0\}$ is a principal ideal, $I≠\{0\}$, let $x_0∈I_0∖\{ 0\}$. Suppose we have non-principal ideals $I_0⊂…⊂I_n$ with non-zero $x_i∈I_i∖\{0\}$ with $x_i|x_{i-1}$ and $x_i≁x_{i-1}$. Since $R$ is a principal ideal in $R$, $I_n≠R$, by Krull's Theorem ∃ maximal ideal $⟨d_n⟩$ such that $I_n⊂⟨d_n⟩$, define $f_n(x)=d_nx$, then $∀x∈I_n:x∈⟨d_n⟩⇒f_n^{-1}(x)≠∅$, also $f_n$ is 1-1 since

Then $I_{n+1}$ is an ideal containing $I_n$ [since $∀x∈I_n:x∈f^{-1}(d_nx)$ where $d_nx∈I_n$].

Since $x_n∈⟨d_n⟩$, $∃x_{n+1}∈I_{n+1}$ such that $x_n=x_{n+1}d_n$

$I_{n+1}$ not principal, [suppose $I_{n+1}=⟨w⟩$, then $I_n=d_n(I_{n+1})=⟨d_nw⟩$ is principal]

By induction on